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I am trying to solve a problem in harmonic functions in Rudin's book(Real and Complex analysis 3rd edition)

To clarify the problem I want to ask, we need some notations:

(1) $U$ is the open unit disc, and $T$ is the unit circle, the boundary of $U$ in the complex plane

(2) $P(z,e^{it})$ is the Poisson Kernel. $$P(z,e^{it})=\frac{1-|z|^2}{|e^{it}-z|^2}$$ for $z\in U$, $e^{it}\in T$

(3)$P[f]$ is the Poisson Integral against $f\in L^1(T)$

(4)$P[d\mu]$ is the Poisson Integral against a complex measure on $T$, defined by $$P[d\mu](z)=\int_T P(z,e^{it})d\mu(e^{it})\quad (z\in U)$$

(5)$C(T)$ is the space consisting of all the continuous complex functions on $T$

(6)We associate to any function $u$ in $U$ a family of functions $u_r$ on $T$, defined by $$u_r(e^{it})=u(re^{it})\quad(0\leq r<1)$$

(7)The measure $\sigma$ is defined by $\sigma=m/2\pi$, where $m$ is ordinary Lebesgue measure on $T$

(8)$||u_r||_1$ is defined by $$||u_r||_1=\int_T |u_r|d\sigma\quad(0\leq r<1)$$

The problem is:

Suppose $u$ is harmonic in $U$, and $\{u_r:0\leq r<1\}$ is a uniformly integrable subset of $L^1(T)$. Modify the proof of Theorem 11.30 to show that $u=P[f]$ for some $f\in L^1(T)$.

Before stating Theorem 11.30, one needs theorem 11.29.

Theorem 11.29: Suppose that (a)$X$ is a separable Banach space, (b)${\Lambda_n}$ is a sequence of linear functionals on $X$, (c)$sup_n||\Lambda_n||=M<\infty$

Then there is a subsequence $\{\Lambda_{n_i}\}$ such that the limit $$\Lambda x=\lim_{i\to\infty}\Lambda_{n_i} x$$ exists for every $x\in X$. Moreover, $\Lambda$ is linear, and $||\Lambda||\leq M$

Proof (Sketch): Note that $\{\Lambda_n\}$ is pointwise bounded and equicontinuous. Since each point of $X$ is a compact set, Theorem 11.29 follows from Arzela-Ascoli Theorem. Besides, it is obvious that $||\Lambda||\leq M$ and that $\Lambda$ is linear.

Theorem 11.30: Suppose $u$ is harmonic in $U$, and $$sup_{0<r<1} ||u_r||_1=M<\infty$$ It follows that there is a unique complex Borel measure $\mu$ on $T$ so that $u=P[d\mu]$

Proof: Define linear functionals $\Lambda_r$ on $C(T)$ by $$\Lambda_r g=\int_T gu_rd\sigma\quad (0\leq r<1)$$ Therefore, $||\Lambda_r||\leq M$. By Theorem 11.29 and Riesz representation theorem for the dual of $C(T)$ there is a measure $\mu$ on $T$, with $||\mu$$||\leq M$, and a sequence $r_j\to 1$, so that $$\lim_{j\to\infty}\int_T gu_{r_j}d\sigma=\int_T gd\mu\quad (*)$$ for every $g\in C(T)$.

Put $h_j(z)=u(r_j z)$. Then $h_j$ is harmonic in $U$, continuous on $\bar{U}$, and is therefore the Poisson integral of its restriction to $T$. Fix $z\in U$, and apply $(*)$ with $$g(e^{it})=P(z,e^{it})$$ Since $h_j(e^{it})=u_{r_j}(e^{it})$, we obtain $$u(z)=\lim_j u(r_j z)=\lim_j h_j(z)$$, and $$\lim_j h_j(z)=\lim_j\int_T P(z,e^{it})h_j(e^{it})d\sigma(e^{it})=\int_T P(z,e^{it})d\mu(e^{it})=P[d\mu](z)$$

To prove uniqueness, it suffices to show that $P[d\mu]=0$ implies $\mu=0$.

Pick $f\in C(T)$, put $u=P[f]$, $v=p[d\mu]$. By Fubini's theorem, and the symmetry $P(re^{i\theta},e^{it})=P(re^{it},e^{i\theta})$, $$\int_T u_rd\mu=\int_T v_rfd\sigma\quad (0\leq r<1)$$ When $v=0$ then $v_r=0$, and since $u_r\to f$ uniformly, as $r\to 1$, we conclude that $$\int_T fd\mu=0$$ for every $f\in C(T)$ if $P[d\mu]=0$. By Riesz representation theorem, $\mu=0$.

Any hints will be appreciated. I've really no idea how to modify the proof of Themorem 11.30, because the $L^1$-boundedness of the family $\{u_r\}$ seemes to play an important role in the proof. However, I cannot see the relationship between the boundedness and uniformly integrability. I have goolged this problem, but I cannot find anything helpful.

Again, any hints will be appreciated.

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up vote 2 down vote accepted

First,Theorem 11.30 draw to showing that

$$\sup_{0<r<1} ||u_r||_1=M<\infty$$

In fact,for a r meet the conditions above.Due to $u_r$ is a harmonic function,the sets $E_1=u_r^{+-1}(\mathbb{R})$ and $E_2=u_r^{--1}(\mathbb{R})$ are measureable.where $u_r=u_r^{+}-u_r^{-}$ .

split $E_j$ into $E_{ji},\mu(E_{ji})<\delta(j=1,2)$ and use the definition of uniformly integrable we can get there exist $M$ such that $sup_{0<r<1} ||u_r||_1=M<\infty$ for all r.

Second,use the Theorem 11.30,we get a Borel measure $\mu$ and $u=P[\mu]$. Then by Lebesgue decomposition,$d\mu = fd\sigma+d\mu_s$ where $f\in L^1(\mu)$ .

Finally,use the definition of uniformly integrable again, it's clear that $\mu_s=0$.

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